Integration of $e^{ax}cos^n bx$ and $e^{ax}sin^n bx$












0















Q: Integration of $e^{ax}cos^n bx$ and $e^{ax}sin^n bx$




I know how to integration $e^{ax}cos bx$ using $cos bx=frac{e^{ibx}+e^{-ibx}}{2}$.Using the same trick here I got $$int e^{ax}cos^n bx ~dx=int e^{ax}left(frac{e^{ibx}+e^{-ibx}}{2}right)^n~ dx$$But it doesn't look easy to solve.I google it and find the result only which is:




$$int e^{ax}cos^n bx ~dx=frac{bnsin (bx)+acos (bx)}{a^2+b^2n^2}e^{ax}+frac{b^2(n-1)nint e^{ax}cos^{n-2}bxdx}{a^2+b^2n^2}.$$




Can Anyone help me to figure out this. Any hints or solution will be appreciated.

Thanks in advance.






complex-integration reduction-formula







share|cite|improve this question
























  • That "result" is a reduction formula, proved by integrating by parts.
    – Lord Shark the Unknown
    Nov 24 at 6:16










  • You can expand your formula by the binomial theorem and integrate termwise.
    – Lord Shark the Unknown
    Nov 24 at 6:17










  • If i use binomial theorem then it will consume a lot of time.Is there any other way to figure out this @Lord Shark the Unknown Sir
    – emonhossain
    Nov 24 at 6:18










  • As alreasy said, integrate by parts to get the reduction formula.
    – Claude Leibovici
    Nov 24 at 7:00










  • You may look at the method I used here.
    – Tianlalu
    Nov 24 at 7:09
















0















Q: Integration of $e^{ax}cos^n bx$ and $e^{ax}sin^n bx$




I know how to integration $e^{ax}cos bx$ using $cos bx=frac{e^{ibx}+e^{-ibx}}{2}$.Using the same trick here I got $$int e^{ax}cos^n bx ~dx=int e^{ax}left(frac{e^{ibx}+e^{-ibx}}{2}right)^n~ dx$$But it doesn't look easy to solve.I google it and find the result only which is:




$$int e^{ax}cos^n bx ~dx=frac{bnsin (bx)+acos (bx)}{a^2+b^2n^2}e^{ax}+frac{b^2(n-1)nint e^{ax}cos^{n-2}bxdx}{a^2+b^2n^2}.$$




Can Anyone help me to figure out this. Any hints or solution will be appreciated.

Thanks in advance.






complex-integration reduction-formula







share|cite|improve this question
























  • That "result" is a reduction formula, proved by integrating by parts.
    – Lord Shark the Unknown
    Nov 24 at 6:16










  • You can expand your formula by the binomial theorem and integrate termwise.
    – Lord Shark the Unknown
    Nov 24 at 6:17










  • If i use binomial theorem then it will consume a lot of time.Is there any other way to figure out this @Lord Shark the Unknown Sir
    – emonhossain
    Nov 24 at 6:18










  • As alreasy said, integrate by parts to get the reduction formula.
    – Claude Leibovici
    Nov 24 at 7:00










  • You may look at the method I used here.
    – Tianlalu
    Nov 24 at 7:09














0












0








0


2






Q: Integration of $e^{ax}cos^n bx$ and $e^{ax}sin^n bx$




I know how to integration $e^{ax}cos bx$ using $cos bx=frac{e^{ibx}+e^{-ibx}}{2}$.Using the same trick here I got $$int e^{ax}cos^n bx ~dx=int e^{ax}left(frac{e^{ibx}+e^{-ibx}}{2}right)^n~ dx$$But it doesn't look easy to solve.I google it and find the result only which is:




$$int e^{ax}cos^n bx ~dx=frac{bnsin (bx)+acos (bx)}{a^2+b^2n^2}e^{ax}+frac{b^2(n-1)nint e^{ax}cos^{n-2}bxdx}{a^2+b^2n^2}.$$




Can Anyone help me to figure out this. Any hints or solution will be appreciated.

Thanks in advance.






complex-integration reduction-formula







share|cite|improve this question
















Q: Integration of $e^{ax}cos^n bx$ and $e^{ax}sin^n bx$




I know how to integration $e^{ax}cos bx$ using $cos bx=frac{e^{ibx}+e^{-ibx}}{2}$.Using the same trick here I got $$int e^{ax}cos^n bx ~dx=int e^{ax}left(frac{e^{ibx}+e^{-ibx}}{2}right)^n~ dx$$But it doesn't look easy to solve.I google it and find the result only which is:




$$int e^{ax}cos^n bx ~dx=frac{bnsin (bx)+acos (bx)}{a^2+b^2n^2}e^{ax}+frac{b^2(n-1)nint e^{ax}cos^{n-2}bxdx}{a^2+b^2n^2}.$$




Can Anyone help me to figure out this. Any hints or solution will be appreciated.

Thanks in advance.






complex-integration reduction-formula




complex-integration reduction-formula






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 4 at 4:19









clathratus

3,017330




3,017330










asked Nov 24 at 6:14









emonhossain

33611




33611












  • That "result" is a reduction formula, proved by integrating by parts.
    – Lord Shark the Unknown
    Nov 24 at 6:16










  • You can expand your formula by the binomial theorem and integrate termwise.
    – Lord Shark the Unknown
    Nov 24 at 6:17










  • If i use binomial theorem then it will consume a lot of time.Is there any other way to figure out this @Lord Shark the Unknown Sir
    – emonhossain
    Nov 24 at 6:18










  • As alreasy said, integrate by parts to get the reduction formula.
    – Claude Leibovici
    Nov 24 at 7:00










  • You may look at the method I used here.
    – Tianlalu
    Nov 24 at 7:09


















  • That "result" is a reduction formula, proved by integrating by parts.
    – Lord Shark the Unknown
    Nov 24 at 6:16










  • You can expand your formula by the binomial theorem and integrate termwise.
    – Lord Shark the Unknown
    Nov 24 at 6:17










  • If i use binomial theorem then it will consume a lot of time.Is there any other way to figure out this @Lord Shark the Unknown Sir
    – emonhossain
    Nov 24 at 6:18










  • As alreasy said, integrate by parts to get the reduction formula.
    – Claude Leibovici
    Nov 24 at 7:00










  • You may look at the method I used here.
    – Tianlalu
    Nov 24 at 7:09
















That "result" is a reduction formula, proved by integrating by parts.
– Lord Shark the Unknown
Nov 24 at 6:16




That "result" is a reduction formula, proved by integrating by parts.
– Lord Shark the Unknown
Nov 24 at 6:16












You can expand your formula by the binomial theorem and integrate termwise.
– Lord Shark the Unknown
Nov 24 at 6:17




You can expand your formula by the binomial theorem and integrate termwise.
– Lord Shark the Unknown
Nov 24 at 6:17












If i use binomial theorem then it will consume a lot of time.Is there any other way to figure out this @Lord Shark the Unknown Sir
– emonhossain
Nov 24 at 6:18




If i use binomial theorem then it will consume a lot of time.Is there any other way to figure out this @Lord Shark the Unknown Sir
– emonhossain
Nov 24 at 6:18












As alreasy said, integrate by parts to get the reduction formula.
– Claude Leibovici
Nov 24 at 7:00




As alreasy said, integrate by parts to get the reduction formula.
– Claude Leibovici
Nov 24 at 7:00












You may look at the method I used here.
– Tianlalu
Nov 24 at 7:09




You may look at the method I used here.
– Tianlalu
Nov 24 at 7:09










1 Answer
1






active

oldest

votes


















1














(Differentiation is easier than integration)



Let $f_{m,a, b}(x)=f_m=e^{ax}cos^m bx$. Take the derivative of $f_{m}$,
begin{align*}
f'_{m}&=af_{m}-bmf_{m-1}sin bx.tag{1}
end{align*}
So, if we take $g_{m}=af_{m}color{magenta}+bmf_{m-1}sin bx$, the derivative of $g_{m}$ will cancel $color{blue}{sin bx}$ out, i.e.
begin{align*}
g'_{m}&=af'_{m}hspace{8.2em}+bmbig(f'_{m-1}sin bxhspace{10.8em}+bunderbrace{f_{m-1}cos bx}_{=f_{m}}big)\
g'_{m}&=a^2f_{m}-color{blue}{abmf_{m-1}sin bx}+color{blue}{bm}big(color{blue}{af_{m-1}sin bx}-b(m-1) f_{m-2}color{orange}{sin^2 bx}+bf_mbig)\
g'_{m}&=(a^2+b^2m)f_m-b^2m(m-1)f_{m-2}color{orange}{(1-cos^2 bx)}\
g'_{m}&=(a^2+b^2m^2)f_m-b^2m(m-1)f_{m-2}.tag{2}
end{align*}
Now, we assume
$$I_{m}=int f_m~mathrm dx=int e^{ax}cos^m bx~mathrm dx,$$
integrate $(2)$, we get the desired result




$$g_m=(a^2+b^2m^2)I_{m}-b^2m(m-1)I_{m-2}.$$







share|cite|improve this answer





















  • Thanks @Tianlalu Really appreciated.Can i use Differentiation to get Integration in this way always?
    – emonhossain
    Nov 24 at 9:05






  • 1




    Yes, we can always do so once we find the appropriate $f_m$. For example, for $int cos^n x~dx$, we may choose $f_n=cos^{n-1}x sin x$ and differentiate it to cancel $sin x$ out.
    – Tianlalu
    Nov 24 at 9:11













Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3011253%2fintegration-of-eax-cosn-bx-and-eax-sinn-bx%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














(Differentiation is easier than integration)



Let $f_{m,a, b}(x)=f_m=e^{ax}cos^m bx$. Take the derivative of $f_{m}$,
begin{align*}
f'_{m}&=af_{m}-bmf_{m-1}sin bx.tag{1}
end{align*}
So, if we take $g_{m}=af_{m}color{magenta}+bmf_{m-1}sin bx$, the derivative of $g_{m}$ will cancel $color{blue}{sin bx}$ out, i.e.
begin{align*}
g'_{m}&=af'_{m}hspace{8.2em}+bmbig(f'_{m-1}sin bxhspace{10.8em}+bunderbrace{f_{m-1}cos bx}_{=f_{m}}big)\
g'_{m}&=a^2f_{m}-color{blue}{abmf_{m-1}sin bx}+color{blue}{bm}big(color{blue}{af_{m-1}sin bx}-b(m-1) f_{m-2}color{orange}{sin^2 bx}+bf_mbig)\
g'_{m}&=(a^2+b^2m)f_m-b^2m(m-1)f_{m-2}color{orange}{(1-cos^2 bx)}\
g'_{m}&=(a^2+b^2m^2)f_m-b^2m(m-1)f_{m-2}.tag{2}
end{align*}
Now, we assume
$$I_{m}=int f_m~mathrm dx=int e^{ax}cos^m bx~mathrm dx,$$
integrate $(2)$, we get the desired result




$$g_m=(a^2+b^2m^2)I_{m}-b^2m(m-1)I_{m-2}.$$







share|cite|improve this answer





















  • Thanks @Tianlalu Really appreciated.Can i use Differentiation to get Integration in this way always?
    – emonhossain
    Nov 24 at 9:05






  • 1




    Yes, we can always do so once we find the appropriate $f_m$. For example, for $int cos^n x~dx$, we may choose $f_n=cos^{n-1}x sin x$ and differentiate it to cancel $sin x$ out.
    – Tianlalu
    Nov 24 at 9:11


















1














(Differentiation is easier than integration)



Let $f_{m,a, b}(x)=f_m=e^{ax}cos^m bx$. Take the derivative of $f_{m}$,
begin{align*}
f'_{m}&=af_{m}-bmf_{m-1}sin bx.tag{1}
end{align*}
So, if we take $g_{m}=af_{m}color{magenta}+bmf_{m-1}sin bx$, the derivative of $g_{m}$ will cancel $color{blue}{sin bx}$ out, i.e.
begin{align*}
g'_{m}&=af'_{m}hspace{8.2em}+bmbig(f'_{m-1}sin bxhspace{10.8em}+bunderbrace{f_{m-1}cos bx}_{=f_{m}}big)\
g'_{m}&=a^2f_{m}-color{blue}{abmf_{m-1}sin bx}+color{blue}{bm}big(color{blue}{af_{m-1}sin bx}-b(m-1) f_{m-2}color{orange}{sin^2 bx}+bf_mbig)\
g'_{m}&=(a^2+b^2m)f_m-b^2m(m-1)f_{m-2}color{orange}{(1-cos^2 bx)}\
g'_{m}&=(a^2+b^2m^2)f_m-b^2m(m-1)f_{m-2}.tag{2}
end{align*}
Now, we assume
$$I_{m}=int f_m~mathrm dx=int e^{ax}cos^m bx~mathrm dx,$$
integrate $(2)$, we get the desired result




$$g_m=(a^2+b^2m^2)I_{m}-b^2m(m-1)I_{m-2}.$$







share|cite|improve this answer





















  • Thanks @Tianlalu Really appreciated.Can i use Differentiation to get Integration in this way always?
    – emonhossain
    Nov 24 at 9:05






  • 1




    Yes, we can always do so once we find the appropriate $f_m$. For example, for $int cos^n x~dx$, we may choose $f_n=cos^{n-1}x sin x$ and differentiate it to cancel $sin x$ out.
    – Tianlalu
    Nov 24 at 9:11
















1












1








1






(Differentiation is easier than integration)



Let $f_{m,a, b}(x)=f_m=e^{ax}cos^m bx$. Take the derivative of $f_{m}$,
begin{align*}
f'_{m}&=af_{m}-bmf_{m-1}sin bx.tag{1}
end{align*}
So, if we take $g_{m}=af_{m}color{magenta}+bmf_{m-1}sin bx$, the derivative of $g_{m}$ will cancel $color{blue}{sin bx}$ out, i.e.
begin{align*}
g'_{m}&=af'_{m}hspace{8.2em}+bmbig(f'_{m-1}sin bxhspace{10.8em}+bunderbrace{f_{m-1}cos bx}_{=f_{m}}big)\
g'_{m}&=a^2f_{m}-color{blue}{abmf_{m-1}sin bx}+color{blue}{bm}big(color{blue}{af_{m-1}sin bx}-b(m-1) f_{m-2}color{orange}{sin^2 bx}+bf_mbig)\
g'_{m}&=(a^2+b^2m)f_m-b^2m(m-1)f_{m-2}color{orange}{(1-cos^2 bx)}\
g'_{m}&=(a^2+b^2m^2)f_m-b^2m(m-1)f_{m-2}.tag{2}
end{align*}
Now, we assume
$$I_{m}=int f_m~mathrm dx=int e^{ax}cos^m bx~mathrm dx,$$
integrate $(2)$, we get the desired result




$$g_m=(a^2+b^2m^2)I_{m}-b^2m(m-1)I_{m-2}.$$







share|cite|improve this answer












(Differentiation is easier than integration)



Let $f_{m,a, b}(x)=f_m=e^{ax}cos^m bx$. Take the derivative of $f_{m}$,
begin{align*}
f'_{m}&=af_{m}-bmf_{m-1}sin bx.tag{1}
end{align*}
So, if we take $g_{m}=af_{m}color{magenta}+bmf_{m-1}sin bx$, the derivative of $g_{m}$ will cancel $color{blue}{sin bx}$ out, i.e.
begin{align*}
g'_{m}&=af'_{m}hspace{8.2em}+bmbig(f'_{m-1}sin bxhspace{10.8em}+bunderbrace{f_{m-1}cos bx}_{=f_{m}}big)\
g'_{m}&=a^2f_{m}-color{blue}{abmf_{m-1}sin bx}+color{blue}{bm}big(color{blue}{af_{m-1}sin bx}-b(m-1) f_{m-2}color{orange}{sin^2 bx}+bf_mbig)\
g'_{m}&=(a^2+b^2m)f_m-b^2m(m-1)f_{m-2}color{orange}{(1-cos^2 bx)}\
g'_{m}&=(a^2+b^2m^2)f_m-b^2m(m-1)f_{m-2}.tag{2}
end{align*}
Now, we assume
$$I_{m}=int f_m~mathrm dx=int e^{ax}cos^m bx~mathrm dx,$$
integrate $(2)$, we get the desired result




$$g_m=(a^2+b^2m^2)I_{m}-b^2m(m-1)I_{m-2}.$$








share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 24 at 8:42









Tianlalu

3,12321038




3,12321038












  • Thanks @Tianlalu Really appreciated.Can i use Differentiation to get Integration in this way always?
    – emonhossain
    Nov 24 at 9:05






  • 1




    Yes, we can always do so once we find the appropriate $f_m$. For example, for $int cos^n x~dx$, we may choose $f_n=cos^{n-1}x sin x$ and differentiate it to cancel $sin x$ out.
    – Tianlalu
    Nov 24 at 9:11




















  • Thanks @Tianlalu Really appreciated.Can i use Differentiation to get Integration in this way always?
    – emonhossain
    Nov 24 at 9:05






  • 1




    Yes, we can always do so once we find the appropriate $f_m$. For example, for $int cos^n x~dx$, we may choose $f_n=cos^{n-1}x sin x$ and differentiate it to cancel $sin x$ out.
    – Tianlalu
    Nov 24 at 9:11


















Thanks @Tianlalu Really appreciated.Can i use Differentiation to get Integration in this way always?
– emonhossain
Nov 24 at 9:05




Thanks @Tianlalu Really appreciated.Can i use Differentiation to get Integration in this way always?
– emonhossain
Nov 24 at 9:05




1




1




Yes, we can always do so once we find the appropriate $f_m$. For example, for $int cos^n x~dx$, we may choose $f_n=cos^{n-1}x sin x$ and differentiate it to cancel $sin x$ out.
– Tianlalu
Nov 24 at 9:11






Yes, we can always do so once we find the appropriate $f_m$. For example, for $int cos^n x~dx$, we may choose $f_n=cos^{n-1}x sin x$ and differentiate it to cancel $sin x$ out.
– Tianlalu
Nov 24 at 9:11




















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3011253%2fintegration-of-eax-cosn-bx-and-eax-sinn-bx%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Plaza Victoria

Image
Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
Read more

In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

Image
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 OneNote has two useful keyboard shortcuts for bulleted and numbered lists, namely Ctrl + . and Ctrl + - . Is there something similar for PowerPoint where the work with lists is also very common? Haven't found anything here: Use keyboard shortcuts in PowerPoint. keyboard-shortcuts microsoft-powerpoint share | improve this question edited Aug 14 '14 at 14:49 Linger 2,826 10 28 40
Read more

How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

Image
1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
Read more